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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2009

  • question_answer
    The length of the diameter of the circle which  cuts three circles \[{{x}^{2}}+{{y}^{2}}-x-y-14=0\] \[{{x}^{2}}+{{y}^{2}}+3x-5y-10=0\] \[{{x}^{2}}+{{y}^{2}}-2x+3y-27=0\] orthogonally, is

    A)  \[8\]                                    

    B)  \[6\]

    C)  \[4\]                                    

    D)  \[2\]

    Correct Answer: C

    Solution :

    Let the equation of circle be \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]     ?..                 and centre \[(-g,-f)\]                 Centres of given circles are                 \[{{C}_{1}}\left( \frac{1}{2},\frac{1}{2} \right),\] \[{{C}_{2}}\left( -\frac{3}{2},\frac{5}{2} \right),\]\[{{C}_{3}}\left( 1,-\frac{3}{2} \right)\] Since, the Eq.  cut the given circles orthogonally. \[\therefore \]  \[-g-f=c=14\]                     ?..(i)                 \[3g-5f=c-10\]   ?..(ii) and        \[-2g+3f=c-27\] ?...(iii) On solving Eqs. (i), (ii) and (iii), we get                 \[g=-3,\] \[f=-4,\] \[c=21\] \[\therefore \]  From Eq. ,  the circle is                 \[{{x}^{2}}+{{y}^{2}}-6x-8y+21=0\] \[\therefore \] Length of diameter                 \[=2\sqrt{{{(-3)}^{2}}+{{(-4)}^{2}}-21}=4\]


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